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4. Functions of Random Variables and Sampling Distribution 221

away from the structure of the multivariate normality for (X , ..., X ), the
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marginals of X , ..., X can each be normal in a particular case, but their sum
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need not be distributed as a normal variable. We recall an earlier example.
Example 4.7.1 (Example 3.6.3 Continued) Let us rewrite the bivariate
normal pdf given in (3.6.1) as . Next, consider any
arbitrary 0 < α, ρ < 1 and fix them. We recall that we had

for –∞ < x , x ∞. Here, (X , X ) has a bivariate non-normal distribution, but
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marginally both X , X have the standard normal distribution. However one
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can show that X + X is not a univariate normal variable. These claims are
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left as the Exercises 4.7.1.-4.7.2. !
4.7.2 Reproductivity of Chi-square Distributions
Suppose that X and X are independent random variables which are respec-
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tively distributed as and . From the Theorem 4.3.2, part (iii), we can
then claim that X + X will be distributed as . But, will the same con-
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clusion necessarily hold if X and X are respectively distributed as and
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, but X and X are dependent? The following simple example shows that
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when X and X are dependent, but X and X are both Chi-squares, then X +
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X need not be Chi-square.
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Example 4.7.2 Suppose that (U , V ) is distributed as N (0, 0, 1, 1, ρ).
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We assume that –1 < ρ < 1. Let and thus the marginal
distributions of X , X are both . Let us investigate the distribution of Y =
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X + X . Observe that
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where the joint distribution of (U + V , U – V ) is actually N , because any
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linear function of U + V , U – V is ultimately a linear function of U and V ,
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and is thus univariate normal. This follows (Definition 4.6.1) from the fact
that (U , V ) is assumed N . Now, we can write
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and hence from our earlier discussions (Theorem 3.7.1), it follows that
U + V and U – V are independent random variables. Also, U + V and
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U – V are respectively distributed as N (0, 2(1 + ρ)) and N (0, 2(1 – ρ)).
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